F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum Mechanics

Einstein's Photoelectric Equation: Solving Problems

F2.2 F3.1 F2.1About 70 minutes

Learning goals

I can use Einstein's photoelectric equation, E(k,max) = hf − W, to solve problems.
I can find the threshold frequency and threshold wavelength of a metal from its work function.
I can interpret a graph of E(k,max) against frequency: the slope is h, the horizontal intercept is f₀ and the vertical intercept is −W.
I can use the law of conservation of energy to explain the photoelectric effect.

Words to know

photoelectric equation
E(k,max) = hf − W: the maximum kinetic energy of photoelectrons equals the photon energy minus the work function.

work function (W)
The minimum energy needed to free an electron from a metal surface; W = hf₀.

threshold frequency (f₀)
The lowest frequency that ejects electrons: f₀ = W/h.

threshold wavelength (λ₀)
The longest wavelength that ejects electrons: λ₀ = hc/W.

conservation of energy
Energy is neither created nor destroyed; in the photoelectric effect, photon energy = work function + kinetic energy.

Energy bookkeeping

Einstein's explanation is really just conservation of energy applied to one photon and one electron:

photon energy = energy to escape + kinetic energy left over

hf = W + E(k,max), usually written E(k,max) = hf − W

The electrons that escape with the most kinetic energy are the ones that needed exactly W to get out. Electrons deeper in the metal lose a little extra energy on the way, so they come out slower. That is why we always talk about the maximum kinetic energy.

Related formulas:

At threshold, E(k,max) = 0, so W = hf₀, and f₀ = W/h.
The threshold (longest) wavelength is λ₀ = c/f₀ = hc/W. In eV and nm: λ₀ = 1240 ÷ W.
From the last lesson: E(k,max) = eV₀.

Energy bookkeeping for one photon and one electron. The photon's energy hf first pays the work function W; whatever is left becomes the kinetic energy of the fastest electrons.

Metal	Work function W (eV)	Threshold wavelength λ₀ (nm)
cesium	2.1	590
sodium	2.3	540
potassium	2.3	540
calcium	2.9	430
aluminium	4.1	300
zinc	4.3	290
copper	4.7	260
platinum	6.4	190

Typical work functions. Measured values vary by a few tenths of an eV with how clean and smooth the surface is. The first three respond to most visible light; calcium responds only to violet light.

Worked example

Example 1: Sodium in violet light

Violet light of wavelength 400 nm shines on sodium (W = 2.3 eV). Find (a) the photon energy, (b) E(k,max) of the photoelectrons, and (c) the stopping voltage.

(a) E = hc/λ = 1240 eV·nm ÷ 400 nm = 3.10 eV.
(b) E(k,max) = hf − W = 3.10 eV − 2.3 eV = 0.80 eV (= 1.3 × 10⁻¹⁹ J).
(c) E(k,max) = eV₀, so V₀ = 0.80 V.

(a) 3.10 eV (b) 0.80 eV (c) 0.80 V

Worked example

Example 2: Why zinc needs ultraviolet

Find the threshold frequency and threshold wavelength for zinc (W = 4.3 eV). Will visible light eject electrons from zinc?

Convert W to joules: 4.3 × 1.60 × 10⁻¹⁹ = 6.88 × 10⁻¹⁹ J.
f₀ = W/h = 6.88 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 1.04 × 10¹⁵ Hz.
λ₀ = c/f₀ = 3.00 × 10⁸ ÷ 1.04 × 10¹⁵ = 2.89 × 10⁻⁷ m = 289 nm. (Shortcut: 1240 ÷ 4.3 = 288 nm.)
Visible light has λ from 400 nm to 700 nm, all longer than 289 nm, so each visible photon has less than 4.3 eV.

f₀ ≈ 1.0 × 10¹⁵ Hz, λ₀ ≈ 290 nm (ultraviolet). No visible light can eject electrons from zinc, which is why Hertz and others needed UV.

Worked example

Example 3: Identifying a metal from data

Ultraviolet light of wavelength 254 nm (from a mercury lamp) falls on an unknown metal. The stopping voltage is 0.60 V. Find the work function and use the table to suggest the metal.

Photon energy: E = 1240 ÷ 254 = 4.88 eV.
E(k,max) = eV₀ = 0.60 eV.
W = hf − E(k,max) = 4.88 − 0.60 = 4.28 eV.

W ≈ 4.3 eV, which matches zinc.

The graph that measures h

Rearrange the equation into the form y = mx + b:

E(k,max) = h f − W

y = E(k,max) (measured from the stopping voltage)
x = f (the frequency of the light)
slope m = h, the same for every metal
vertical intercept b = −W
horizontal intercept = f₀

Millikan plotted exactly this graph in 1916. His points fell on a straight line whose slope agreed with Planck's constant, which had been measured in a completely different way (from blackbody spectra). That agreement was powerful evidence for Einstein's photon model.

E(k,max) against frequency for sodium and zinc. The lines are parallel (slope h = 4.14 × 10⁻¹⁵ eV·s), cross the frequency axis at each metal's threshold f₀, and extend back (dotted) to −W on the vertical axis.

Worked example

Example 4: Reading the slope

A student's best-fit line for E(k,max) (in eV) against f passes through (5.0 × 10¹⁴ Hz, 0.41 eV) and (9.0 × 10¹⁴ Hz, 2.07 eV). Find h in J·s.

Slope = Δy/Δx = (2.07 − 0.41) eV ÷ (9.0 × 10¹⁴ − 5.0 × 10¹⁴) Hz = 1.66 eV ÷ 4.0 × 10¹⁴ Hz = 4.15 × 10⁻¹⁵ eV·s.
Convert eV to J: 4.15 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.64 × 10⁻³⁴ J·s.

h ≈ 6.6 × 10⁻³⁴ J·s, very close to the accepted 6.63 × 10⁻³⁴ J·s.

Problem-solving checklist

Decide which unit you will use (eV with 1240 eV·nm, or J with h and c) and stay in it.
Find the photon energy first.
Compare with W. If hf < W, stop: no electrons, and E(k,max) is not negative.
Subtract to get E(k,max); use eV₀ = E(k,max) for the stopping voltage; use ½mv² for the speed (in joules!).

Just for fun

The work function is a metal's cover charge. Every electron pays W at the door, and whatever energy it has left it spends on speed. Photons below the threshold are turned away, however big their group is.

Hands-on activity

Graph photoelectric data (data practice)

You need: paper, a ruler and a pencil (graph paper helps), or a spreadsheet.

A student measured these stopping voltages for one metal:

λ (nm): 300, 350, 400, 450, 500
V₀ (V): 2.03, 1.44, 1.00, 0.66, 0.38

Convert each wavelength to frequency with f = c/λ (keep 3 significant digits).
Each E(k,max) in eV equals V₀. Plot E(k,max) (eV) on the y-axis against f (× 10¹⁴ Hz) on the x-axis.
Draw the best-fit straight line. Find its slope and convert to J·s.
Read the vertical intercept (extend the line) to get −W, and the horizontal intercept to get f₀.
Use the table above to identify the metal.

What to look for: frequencies 10.0, 8.57, 7.50, 6.67 and 6.00 (× 10¹⁴ Hz). The points lie on a line of slope about 4.15 × 10⁻¹⁵ eV·s, which is about 6.6 × 10⁻³⁴ J·s. The intercepts give W ≈ 2.1 eV and f₀ ≈ 5.0 × 10¹⁴ Hz: the metal is cesium.

Your turn

Light of wavelength 500 nm falls on cesium (W = 2.1 eV). Find E(k,max) and the stopping voltage.

Find the threshold wavelength for calcium (W = 2.9 eV).

Light of 300 nm falls on zinc (W = 4.3 eV). Are electrons emitted?

A metal's threshold frequency is 6.0 × 10¹⁴ Hz. Find its work function in J and eV.

Light of 200 nm falls on copper (W = 4.7 eV). Find E(k,max) in eV and the maximum speed of the electrons.

On a graph of E(k,max) against f for potassium, what are the slope and the vertical intercept?

Light that gives a stopping voltage of 1.5 V on sodium (W = 2.3 eV) has what wavelength?

More practice: warm-up

Write Einstein's photoelectric equation.

What is the relationship between work function and threshold frequency?

A photon has 3.0 eV and the work function is 2.0 eV. Find E(k,max).

If E(k,max) = 0.50 eV, what is the stopping voltage?

Which metal in the table has the lowest work function?

More practice: on your own

Light of frequency 8.0 × 10¹⁴ Hz falls on a metal with W = 2.0 eV. Find the stopping voltage.

Two metals, A (W = 2.3 eV) and B (W = 4.1 eV), are lit with 220 nm UV. Which gives faster electrons, and what is the difference in E(k,max)?

The wavelength of light on a metal is halved. Does E(k,max) double? Explain with an example.

A metal ejects electrons only for wavelengths shorter than 540 nm. When 360 nm light is used, what is the stopping voltage?

Why are the lines on the E(k,max) vs f graph for different metals always parallel?

Challenge

When light of 400 nm falls on a metal, V₀ = 1.0 V. When light of 300 nm is used, V₀ = 2.0 V. Use only these data to find h and W (take e = 1.60 × 10⁻¹⁹ C, c = 3.00 × 10⁸ m/s).

A 2.0 mW beam of 250 nm light falls on sodium. If 1 photon in 1000 ejects an electron, find (a) the photocurrent and (b) E(k,max).

Explain why a graph of E(k,max) against wavelength would be a curve, not a line, and why physicists prefer to plot against frequency.

Check your understanding

Quiz

Light of photon energy 5.0 eV falls on a metal with a work function of 3.2 eV. The maximum kinetic energy of the photoelectrons is

On a graph of E(k,max) against frequency, the slope equals

A metal has a work function of 2.5 eV. Its threshold wavelength is about

If light of a frequency below f₀ falls on a metal, E(k,max) is

The work function of a metal is W. Light with photon energy 2W shines on it. The stopping voltage is

Lines for different metals on an E(k,max) vs f graph

For parents and teachers

Big idea (F2.2): E(k,max) = hf − W is conservation of energy for one photon and one electron. Its graph (E(k,max) vs f) is a straight line with slope h for every metal.

How to help: check that your teen decides on units first (eV or J) and finds the photon energy before anything else. The data activity is a good one to do together on a spreadsheet; ask what the slope means physically.

Common difficulties:

Reporting a negative kinetic energy instead of saying "no emission."
Mixing units: subtracting a work function in eV from a photon energy in J.
Using eV in ½mv². Speeds require joules.
Thinking halving the wavelength doubles E(k,max). It doubles hf, not hf − W.